Question

Prove that 1/sec theta-tan theta=sec theta+tan theta

Answer 1

Answer:

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Answer 2

$\frac{1}{sec \theta - tan \theta} \\ \\ = \frac{sec \theta + tan \theta}{(sec \theta - tan \theta)(sec \theta + tan \theta)} \\ \\ = \frac{sec \theta + tan \theta}{ {sec}^{2} \theta - {tan}^{2} \theta} \\ \\ = sec \theta + tan \theta$